Linear Programming - Tutorial -- Section 6.8 --

The book does an excellent job developing and explaining linear programming .   Read this section thoroughly and if necessary many times.  I strongly recommend using the tables to organize your data.

Linear Programming
A Graphical Approach  Maximize the objective function 7x + 4y subject to the following constraints.

The question is :  Find the value for the objective function that is the largest.  In order to find that value,  we need to graph the inequalities.

First, we want to graph the line
3x + 2y = 36
Use x and y intercepts.

• To find x intercept, set y = 0
3x + 2(0) = 36
3x = 36
x = 12,   (12, 0)  is x intercept

• To find y intercept, set x = 0
3(0) + 2y = 36
2y = 36
y = 18,   (0, 18)  is y intercept Next, we want to graph the line
x + 4y = 32
Use x and y intercepts.

• To find x intercept, set y = 0
x + 4(0) = 32
x = 32
(32, 0)  is x intercept

• To find y intercept, set x = 0
0 + 4y = 32
4y = 32
y = 8,   (0, 8)  is y intercept Solve each inequality for y   • Since both inequalities have less than symbols ,< , the feasible solution is below each line.
• x ł 0  , y ł 0 means the x and y values are positive.  To the right of the y axis and above the x axis . (quadrant I)
• Note:  If the inequality has a greater than symbol,  ł  , the feasible solution is above the line. Now that the feasible set is shaded in,  we need to find the corner points.  I count 4 corner points.
The first 3 corner points are obvious.  (0, 0), (0, 8), and (12, 0) . Use the equations to the right to find the other point. Find the remaining corner point by setting the equations equal to each other and solve for x. Multiply both sides by 4 We get... adding 6x and subtracting -32 from both sides 40 = 5x
The ordered pair is (8,6) x = 8, then (8, 6), (0, 0), (0, 8), and (12, 0)   are the four corner points.
Finally,  test each corner point in the objective function.

Maximize the objective function 7x + 4y subject to the following constraints.

 Corner Point Value (0,0) 7(0) + 4(0) = 0 + 0 = 0 (0,8) 7(0) + 4(8) = 0 +32 = 32 (12,0) 7(12) + 4(0) = 84 + 0 = 84 (6,8) 7(6) + 4(8) = 42 + 32 = 74 Maximum  is 84 The maximum value 84 occurs at the point (12,0) An oil company owns 2 refineries. Refinery I produces each day 100 barrels of high grade oil, 200 barrels of medium grade oil, and 300 barrels of low grade oil and costs \$10,000 a day to operate. Refinery II produces each day 200 barrels of high grade oil, 100 barrels of medium grade oil, and 200 barrels of low grade oil and costs \$9,000 a day to operate.

An order is received for 1000 barrels of high grade oil, 1000 barrels of medium grade oil, and 1800 barrels of low grade oil.  How many days should each refinery be operated to fill the order at the least cost?

Solution:
Let's start by making a table.

 Refinery I Refinery II Ordered high grade 100 barrels 200 barrels 1000 barrels medium grade 200 barrels 100 barrels 1000 barrels low grade 300 barrels 200 barrels 1800 barrels Cost \$10,000 \$9,000 ?
• Let x be the number of days refinery I is open.
• Let y be the number of days refinery II is open.

Now, write the inequalities:

 high grade 100x + 200y ł 1000 medium grade 200x + 100y ł 1000 low grade 300x + 200y ł 1800

Write the inequalities in standard form by rewrite solving for y.  This is the first inequality in red with x-int (10,0) and y-int (0,5) The solution lies above the line This is the 2nd inequality in blue with x-int (5,0) and y-int (0,10) The solution lies above the line This is the 3rd inequality in purple with x-int (6,0) and y-int (0,9) The solution lies above the line The feasible solution is blue area.

Next find the corner points.  By looking at the graph and knowing x ł 0 and y ł 0 means the solution is in the first quadrant we get (10,0) and (0,10) as corner points.  We must calculate the remaining corner points by setting the corresponding inequalities equal to each other.

 -.5x + 5 = -1.5x + 9 x = 4, then y = -.5(4) + 5 = 3 (4,3) -2x + 10 = -1.5x + 9 x = 2, then y = -2(2) + 10 = 6 (2,6) Note :  The red and blue line intersection is outside the feasible region.
 Finally, let's check the cost function  10,000x + 9,000y  we want to minimize. (0,10) 10,000(0) + 9,000(10) = 90,000 (4,3) 10,000(4) + 9,000(3) = 67,000 (2,6) 10,000(2) + 9,000(6) = 74,000 (10,0) 10,000(10) + 9,000(0) = 100,000

The minimum cost of \$67,000 is achieved at (4,3).
Open refinery I for 4 days and refinery II for 3 days.

Tutorials and Applets by
Joe McDonald